Hess’s law calculations. 2C(s) + 3H 2 (g) + ½O 2 (g) → C 2 H 5 OH(l) C(s) + O 2 (g) → CO 2 (g) ∆H = –393 kJ mol –1 H 2 (g) + ½O 2 (g) → H 2 O(l) ∆H =

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Hess’s law calculations

2C(s) + 3H 2 (g) + ½O 2 (g) → C 2 H 5 OH(l) C(s) + O 2 (g) → CO 2 (g) ∆H = –393 kJ mol –1 H 2 (g) + ½O 2 (g) → H 2 O(l) ∆H = –285 kJ mol –1 C 2 H 5 OH(l) + 3O 2 (g) → 2CO 2 (g) + 3H 2 O(l) ∆H = –1364 kJ mol –1 Calculate the enthalpy change for this reaction: given: 2 C atoms are required as a reactant. C(s) + O 2 (g) → CO 2 (g) ∆H = –393 kJ mol – × 2 3 H 2 molecules are required as a reactant. H 2 (g) + ½O 2 (g) → H 2 O(l) ∆H = –285 kJ mol – × 3 C 2 H 5 OH is a product. C 2 H 5 OH(l) + 3O 2 (g) 2CO 2 (g) + 3H 2 O(l) → ∆H = 1364 kJ mol –1 + Remember to change the sign of the ∆H when reversing the equation.

2C(s) + 2O 2 (g) → 2CO 2 (g) ∆H = –393 kJ mol –1 × 2 3H 2 (g) + 1½O 2 (g) → 3H 2 O(l) ∆H = –285 kJ mol –1 × 3 2CO 2 (g) + 3H 2 O(l) → C 2 H 5 OH(l) + 3O 2 (g) ∆H = kJ mol –1 2C(s) + 3H 2 (g) + ½O 2 (g) → C 2 H 5 OH(l) ∆H = –277 kJ mol –1 Before adding these three equations, cancel out the terms which appear on both sides of the arrows. 2C(s) + 3H 2 (g) + ½O 2 (g) → C 2 H 5 OH(l) Use the bracket keys on your calculator to add up the ∆H values. It’s very easy to make errors in the exam, so do each sum twice.